Hello, can i get some with #30 a and b, ? don't forget to show your work.

a. dy/dx = -2/3

b. dy/dx = -28

Step-by-step explanation:

One way to do this is to assume that x and y are functions of something else, say "t", then differentiate with respect to that. If we write dx/dt = x' and dy/dt = y', then the required derivative is y'/x' = dy/dx.

a. x'·y^3 +x·(3y^2·y') = 0

y'/x' = -y^3/(3xy^2) = -y/(3x)

For the given point, this is ...

dy/dx = -2/3

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b. 2x·x' +x^2·y' -2x'·y^3 -2x·(3y^2·y') + 0 = 2x' + 2y'

y'(x^2 -6xy^2 -2) = x'(2 -2x +2y^3)

y'/x' = 2(1 -x +y^3)/(x^2 +6xy^2 -2)

For the given point, this is ...

dy/dx = 2(1 -0 +27)/(0 +0 -2)

dy/dx = -28

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The attached graphs show these to be plausible values for the derivatives at the given points.

a.  

Step-by-step explanation:

b.  -28

Step-by-step explanation:

idc

step-by-step explanation:

idk